Is it correct to say $\text{max}(x_1,x_2,\dots,x_n)=\lim_{p \to \infty} \frac{x_1^p+x_2^p+\dots+x_n^p}{x_1^{p-1}+x_2^{p-1}+\dots+x_n^{p-1}}$ There are many possible way to represent the maximum function, I came up with this one:
$$\text{max}(x_1,x_2,\dots,x_n)=\lim_{p \to \infty} \frac{x_1^p+x_2^p+\dots+x_n^p}{x_1^{p-1}+x_2^{p-1}+\dots+x_n^{p-1}}$$
$$x_1,x_2,\dots,x_n \geq 0$$
Numerically seems to be true. I'm not sure if my proof is correct though.
Let the following condition hold (without loss of generality, except for the case when $x_1=x_k$ for some $k>1$, but I'm not going to consider it for now):
$$x_1 > x_2 \geq \dots \geq x_n$$
Now let's divide both numerator and denominator by $x_1^{p-1}$:
$$\frac{x_1^p+x_2^p+\dots+x_n^p}{x_1^{p-1}+x_2^{p-1}+\dots+x_n^{p-1}}=\frac{x_1+\frac{x_2^p}{x_1^{p-1}}+\dots+\frac{x_n^p}{x_1^{p-1}}}{1+\frac{x_2^{p-1}}{x_1^{p-1}}+\dots+\frac{x_n^{p-1}}{x_1^{p-1}}}$$
Now one part of the expression poses no trouble when taking the limit:
$$\lim_{p \to \infty} \frac{x_1}{1+\frac{x_2^{p-1}}{x_1^{p-1}}+\dots+\frac{x_n^{p-1}}{x_1^{p-1}}}=x_1$$
As for the rest of the expression, I'm not so sure:
$$\lim_{p \to \infty} \frac{\frac{x_2^p}{x_1^{p-1}}+\dots+\frac{x_n^p}{x_1^{p-1}}}{1+\frac{x_2^{p-1}}{x_1^{p-1}}+\dots+\frac{x_n^{p-1}}{x_1^{p-1}}}=\lim_{p \to \infty} \frac{x_2^p+\dots+x_n^p}{x_1^{p-1}+x_2^{p-1}+\dots+x_n^{p-1}}$$
It should be equal to $0$ if my conjecture is correct, but I don't see how it works. We can't just take $p-1 \to p$ as $p \to \infty$, can we?
 A: The term $\frac{x_2^p}{x_1^{p-1}}=\left(\frac{x_2}{x_1}\right)^{p-1} x_{2}$.  If $x_2$ is constant (which it is) and $-1<\frac{x_2}{x_1}<1$ then the term: $\left(\frac{x_2}{x_1}\right)^{p-1}$ still goes to zero.
I think the only issue you might have is when some of the $x_i$'s are negative (particularly if $|x_n| > |x_1|$).  If you make the extra assumption that the values are non-negative then this looks ok.
A: Your RHS in the last equation should be like this (I think)$$\lim_{p \to \infty} \frac{\left({x_2\over x_1}\right)^{p-1}x_2+\dots+\left({x_n\over x_1}\right)^{p-1}x_n}{1+\frac{x_2^{p-1}}{x_1^{p-1}}+\dots+\frac{x_n^{p-1}}{x_1^{p-1}}}$$ 
Now each of the $\left({x_i\over x_1}\right)$ is less than $1$, So raised to power $\infty$ will make each of them zero and hence the numerator(each $x_i$ is finite ).  And you are done I think..
A: Let $x_k>0$ for $k=1\ldots n$. Suppose, that $x_1=\ldots =x_M>x_{M+1}\ge \ldots x_{n}$ for some $1\le M \le n$. Then clearly:
$$
\frac{x_1^p+\ldots+x_n^p}{x_1^{p-1}+\ldots+x_n^{p-1}}=
x_1\frac{M+\left(\frac{x_{M+1}}{x_1}\right)^p+
\ldots+\left(\frac{x_{n}}{x_1}\right)^p}{M+\left(\frac{x_{M+1}}{x_1}\right)^{p-1}+
\ldots+\left(\frac{x_{n}}{x_1}\right)^{p-1}} \stackrel{p\to \infty}{\to} x_1
$$
(the fractions $\frac{x_k}{x_1}$ are smaller than $1$ for $k>M$)
The positivity is neccessary: the simple $x=1,-1,\frac{1}{2}$ sequence yields a divergent sequence (for $p\in \mathbb{N}$)
Or, we could consider only special sequences, for example strictly dominant's: $|x_1|>|x_2|\ge\ldots \ge |x_n|$. For these, we get $|x_1|$ as limit, which is not what we expect...
